A box is given a push so that it slides across the floor. How far will it go, given that the coefficient of kinetic friction is 0.17 and the push imparts an initial speed of 4.3 m/s?

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just do the math there is a formula

To find how far the box will go, we need to calculate the stopping distance of the box due to friction. The stopping distance is the distance it takes for the box to stop completely.

To find the stopping distance, we'll use the equation of motion for constant friction:

v^2 = u^2 + 2as

Where:
v = final velocity (0 m/s, since the box stops)
u = initial velocity (4.3 m/s)
a = acceleration (friction)
s = stopping distance (what we're trying to find)

To find the acceleration due to friction, we'll use the formula:

f = u * μ * N

Where:
f = force of friction
u = coefficient of kinetic friction (0.17)
N = normal force

The normal force is the force exerted by the surface on the box, perpendicular to the surface. Since the box is on a horizontal surface, the normal force is equal to the weight of the box, which can be calculated as:

N = m * g

Where:
m = mass of the box
g = acceleration due to gravity (approximately 9.8 m/s^2)

Now, let's go step-by-step to find the stopping distance:

Step 1: Calculate the normal force.
N = m * g

Step 2: Calculate the force of friction.
f = u * N

Step 3: Calculate the acceleration due to friction.
a = f / m

Step 4: Use the equation of motion to find the stopping distance.
v^2 = u^2 + 2as

Since v = 0, we have:
0 = u^2 + 2as

Solving for s, we get:
s = - (u^2) / (2a)

Now, let's calculate the stopping distance using the given values:

Step 1: Calculate the normal force.
N = m * g
Assuming the mass of the box is 1 kg, the normal force is given by:
N = 1 kg * 9.8 m/s^2 = 9.8 N

Step 2: Calculate the force of friction.
f = u * N
f = 0.17 * 9.8 N = 1.666 N

Step 3: Calculate the acceleration due to friction.
a = f / m
a = 1.666 N / 1 kg = 1.666 m/s^2

Step 4: Calculate the stopping distance.
s = - (u^2) / (2a)
s = - (4.3 m/s)^2 / (2 * 1.666 m/s^2)
s = - 18.49 m^2/s^2 / 3.332 m/s^2
s ≈ -5.55 m^2

Since distance cannot be negative, the stopping distance is approximately 5.55 meters.

Therefore, the box will go approximately 5.55 meters before coming to a stop.